Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

APPENDIX D - Trigonometry - D Exercises - Page A 32: 33

Answer

$sin\beta=-\frac{1}{\sqrt 10}=-\frac{\sqrt 10}{10}$ $cos\beta=-\frac{3}{\sqrt 10}=-\frac{3\sqrt 10}{10}$ $tan\beta=\frac{1}{ 3}$ $csc\beta=-\sqrt 10$ $sec \beta=-\frac{\sqrt 10}{3}$

Work Step by Step

Since $cot\beta=3$, we can label the adjacent side as having length 3 and the opposite side as having length 1. The Pythagorean Theorem gives hypotenuse side$=\sqrt {3^{2}+1^{2}}=\sqrt 10$ The other five trigonometric functions are given as follows: $sin\beta=-\frac{1}{\sqrt 10}=-\frac{\sqrt 10}{10}$ $cos\beta=-\frac{3}{\sqrt 10}=-\frac{3\sqrt 10}{10}$ $tan\beta=\frac{1}{ 3}$ $csc\beta=-\sqrt 10$ $sec \beta=-\frac{\sqrt 10}{3}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.